Integral Exercise Level 2 for \(\sqrt{{\sec}^{4}x-1}\)

Exercise:

Integrate, $\int{\sqrt{{{\sec }^{4}}x-1}\,dx}$

Solution: we have ${{\sec }^{4}}x-1={{\left( {{\sec }^{2}}x \right)}^{2}}-1=\left( {{\sec }^{2}}x-1 \right)\left( {{\sec }^{2}}x+1 \right)$

But ${{\sec }^{2}}x-1={{\tan }^{2}}x$ hence ${{\sec }^{4}}x-1={{\tan }^{2}}x\left( {{\sec }^{2}}x+1 \right)$

So $\int{\sqrt{{{\sec }^{4}}x-1}\,dx}=\int{\sqrt{{{\tan }^{2}}x\left( {{\sec }^{2}}x+1 \right)}\,dx}=\int{\tan x\sqrt{{{\sec }^{2}}x+1}\,dx}$

Let $u=\tan x\Leftrightarrow du={{\sec }^{2}}xdx\Leftrightarrow dx=\frac{du}{{{\sec }^{2}}x}$

So $\int{\tan x\sqrt{{{\sec }^{2}}x+1}\,dx}=\int{u\sqrt{{{\tan }^{2}}x+1+1}\,\frac{du}{{{\sec }^{2}}x}}=\int{u\sqrt{{{u}^{2}}+2}\frac{du}{{{u}^{2}}+1}}=\int{\frac{u\sqrt{{{u}^{2}}+2}}{{{u}^{2}}+1}du}$

Let ${{w}^{2}}={{u}^{2}}+2\Rightarrow {{w}^{2}}-1={{u}^{2}}+1$ and $2wdw=2udu\Leftrightarrow wdw=udu$

So $\int{\frac{u\sqrt{{{u}^{2}}+2}}{{{u}^{2}}+1}du}=\int{\frac{{{w}^{2}}}{{{w}^{2}}-1}dw=\int{\frac{{{w}^{2}}+1-1}{{{w}^{2}}-1}dw=\int{dw+\int{\frac{dw}{{{w}^{2}}-1}}}}}$

Observe that $\frac{1}{{{w}^{2}}-1}=\frac{1}{\left( w-1 \right)\left( w+1 \right)}=\frac{A}{w-1}+\frac{B}{w+1}=\frac{A\left( w+1 \right)+B\left( w-1 \right)}{{{w}^{2}}-1}$

So $A+B=0\,\,\,\And \,\,\,\,A-B=1$ thus $A=\frac{1}{2}\,\,\,\And \,\,B=\frac{-1}{2}$

Thus $\int{\frac{dw}{{{w}^{2}}-1}=\frac{1}{2}\int{\frac{dw}{w-1}-\frac{1}{2}\int{\frac{dw}{w+1}=\frac{1}{2}\ln \left| \frac{w-1}{w+1} \right|+c}}}$

So $\int{\frac{{{w}^{2}}}{{{w}^{2}}-1}dw}=w+\frac{1}{2}\ln \left| \frac{w-1}{w+1} \right|+c$  by back word substitution we get

$\int{\frac{u\sqrt{{{u}^{2}}+2}}{{{u}^{2}}+1}du}=\sqrt{{{u}^{2}}+2}+\frac{1}{2}\ln \left| \frac{\sqrt{{{u}^{2}}+2}-1}{\sqrt{{{u}^{2}}+2}+1} \right|+c$

Thus $\int{\sqrt{{{\sec }^{4}}x-1}\,dx}=\sqrt{{{\tan }^{2}}x+2}+\frac{1}{2}\ln \left| \frac{\sqrt{{{\tan }^{2}}x+2}-1}{\sqrt{{{\tan }^{2}}x+2}+1} \right|+c$